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(B) The equation of a straight line at a constant distance $P$ from the origin is $x \cos \alpha + y \sin \alpha = P$. Differentiating with respect to

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$6$

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(B) The equation of a straight line at a constant distance $P$ from the origin is $x \cos \alpha + y \sin \alpha = P$. Differentiating with respect to $x$,we get $\cos \alpha + \frac{dy}{dx} \sin \alpha = 0$,which implies $\frac{dy}{dx} = -\cot \alpha$. From this,we have $\cos \alpha = -\frac{dy/dx}{\sqrt{1+(dy/dx)^2}}$ and $\sin \alpha = \frac{1}{\sqrt{1+(dy/dx)^2}}$. Substituting these into the original equation: $x \left( -\frac{dy/dx}{\sqrt{1+(dy/dx)^2}} \right) + y \left( \frac{1}{\sqrt{1+(dy/dx)^2}} \right) = P$. This simplifies to $y - x \frac{dy}{dx} = P \sqrt{1 + (\frac{dy}{dx})^2}$. Squaring both sides,we get $(y - x \frac{dy}{dx})^2 = P^2 (1 + (\frac{dy}{dx})^2)$. The highest derivative present is $\frac{dy}{dx}$,so the order $l = 1$. The highest power of the derivative $\frac{dy}{dx}$ after clearing radicals is $2$,so the degree $m = 2$. Thus,$l m^2 + l^2 m = (1)(2^2) + (1^2)(2) = 4 + 2 = 6$.

If $l$ and $m$ are the order and degree of the differential equation of all straight lines at a constant distance of $P$ units from the origin,then $l m^2+l^2 m=$

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